Coefficient of $x^3$ in $(1-2x+3x^2-4x^3)^{1/2}$ The question is to find out the coefficient of $x^3$ in the expansion of  $(1-2x+3x^2-4x^3)^{1/2}$
I tried using multinomial theorem but here the exponent is a fraction and I couldn't get how to proceed.Any ideas?
 A: Hint:
The coefficient of $x^3$ in the expansion of  $(1-2x+3x^2-4x^3)^{1/2}$
$=$
the coefficient of $x^3$ in the expansion of  $(1-2x+3x^2-4x^3+\cdots)^{1/2}$
Using Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $,
$$1-2x+3x^2-4x^3+\cdots=(1+x)^{-2}$$
$$(1-2x+3x^2-4x^3+\cdots)^{1/2}=(1+x)^{-1}=1-x+x^2-x^3+\cdots$$
A: Say $$(1-2x+3x^2-4x^3)^{1/2} =a+bx+cx^2+dx^3...$$
then 
$$1-2x+3x^2-4x^3 =(a+bx+cx^2+dx^3...)^2$$
but $$(a+bx+cx^2+dx^3...)^2 = a^2+2abx+(2ac+b^2)x^2+2(ad+bc)x^3+...$$
So $a=\pm 1$. 
If $a=1$ then $b=-1$ and $c=1$ and $d=-1$
A: Expand $(1+u)^{\tfrac12}$ up to order $3$:
$$(1+u)^{\tfrac12}=1+\frac12 u-\frac18u^2+\frac1{16}u^3+o(u^3),$$
and compose with $u=-2x+3x^2-4x^3$:


*

*$u^2=4x^2-12x^3+o(x^3)$,

*$u^3=u^2\cdot u=-8x^3+o(x^3)$.
One finally obtains$$1-x+x^2-x^3+o(x^3).$$
A: It is quite practical to exploit the identity
$$ (1+x)^2 (1-2x+3x^2-4x^3) = 1-5x^4-4x^5 \tag{A}$$
from which
$$\begin{eqnarray*} \sqrt{1-2x+3x^2-4x^3} &=& \frac{\sqrt{1-5x^4-4x^3}}{1+x} \\&=&\left(1+O(x^4)\right)(1-x+x^2-x^3+O(x^4))\tag{B}\end{eqnarray*}$$
and the coefficient of $x^3$ in the RHS of $(B)$ is trivially $\color{red}{-1}$.
